posted by racheal .
An equilibrium mixture of SO2,SO3 and O2 gases is maintained in a 11.5 L flask at a temp at which Kc=55.2 for the rxn shown below. If the number of moles of SO3 in the flask at equilibrium is twice the num of moles of SO2, how much O2 is present in the flask at equilibrium?
2SO2 (g) + O2 --> 2SO3(g)
.........2SO2 + O2 ==> 2SO3
At equilibrium, concns are
(SO2) = (x/11.5) M
(SO3) = (2x/11.5)M
(O2) = y M
Kc = 55.2 = (2x/11.5)^2/(x/11.5)^2*y
Solve for y which gives (O2) in moles/L, then multiply by 11.5L to obtain moles.
You can do it another way by assuming any convenient number for moles SO2 (e.g., 11.5 moles) for SO2 so (SO3) = 11.5/11.5 = 1M
Then moles SO3 = 23.0 moles or (SO3) = 23.0/11.5 = 2.0M
O2 = xM and solve for x, then multiply by 11.5 to find moles O2.